从R中的另一列创建列并为其加电
目标是从下面数据的代码列创建两个新列。一个带数字,另一个带代码(系数)。我该怎么做?我尝试了从R中的另一列创建列并为其加电,r,R,目标是从下面数据的代码列创建两个新列。一个带数字,另一个带代码(系数)。我该怎么做?我尝试了ifelse(),但它出现了错误 structure(list(Potreiro = structure(c(3L, 3L, 3L, 3L, 3L, 4L, 3L, 4L, 3L, 4L), .Label = c("1A", "6B", "7A", "7B"), class = "factor"), Code = structure(c(4L, 1L, 8L, 3L, 2L, 4L, 6L, 5L,
ifelse()
,但它出现了错误
structure(list(Potreiro = structure(c(3L, 3L, 3L, 3L,
3L, 4L, 3L, 4L, 3L, 4L), .Label = c("1A", "6B", "7A", "7B"), class =
"factor"), Code = structure(c(4L, 1L, 8L, 3L, 2L, 4L, 6L, 5L, 8L, 7L ),
.Label = c("2", "3", "4", "5", "50%", "70%", "ac", "ad", "av", "cd", "de",
"Dem"), class = "factor")), .Names = c("Potreiro", "Code"), row.names =
c(NA, 10L), class = "data.frame")
谢谢 我会这样做:
df <-
structure(list(Potreiro = structure(c(3L, 3L, 3L, 3L,
3L, 4L, 3L, 4L, 3L, 4L), .Label = c("1A", "6B", "7A", "7B"), class =
"factor"), Code = structure(c(4L, 1L, 8L, 3L, 2L, 4L, 6L, 5L, 8L, 7L ),
.Label = c("2", "3", "4", "5", "50%", "70%", "ac", "ad", "av", "cd", "de",
"Dem"), class = "factor")), .Names = c("Potreiro", "Code"), row.names =
c(NA, 10L), class = "data.frame")
arenum <- sapply(df$Code, function (x) grepl('[[:digit:]]', x))
df$codenum <- ifelse(arenum, as.character(df$Code), NaN)
df$codechar <- ifelse(!arenum, as.character(df$Code), NaN)
df
df我会这样做:
df <-
structure(list(Potreiro = structure(c(3L, 3L, 3L, 3L,
3L, 4L, 3L, 4L, 3L, 4L), .Label = c("1A", "6B", "7A", "7B"), class =
"factor"), Code = structure(c(4L, 1L, 8L, 3L, 2L, 4L, 6L, 5L, 8L, 7L ),
.Label = c("2", "3", "4", "5", "50%", "70%", "ac", "ad", "av", "cd", "de",
"Dem"), class = "factor")), .Names = c("Potreiro", "Code"), row.names =
c(NA, 10L), class = "data.frame")
arenum <- sapply(df$Code, function (x) grepl('[[:digit:]]', x))
df$codenum <- ifelse(arenum, as.character(df$Code), NaN)
df$codechar <- ifelse(!arenum, as.character(df$Code), NaN)
df
df这里有一个使用extract
library(dplyr)
library(tidyr)
df %>%
extract(Code, into = c('number', 'word'), '(\\d*)([a-z]*)', remove = FALSE, convert = TRUE)
# Potreiro Code number word
#1 7A 5 5
#2 7A 2 2
#3 7A ad NA ad
#4 7A 4 4
#5 7A 3 3
#6 7B 5 5
#7 7A 70% 70
#8 7B 50% 50
#9 7A ad NA ad
#10 7B ac NA ac
这里有一个使用extract
library(dplyr)
library(tidyr)
df %>%
extract(Code, into = c('number', 'word'), '(\\d*)([a-z]*)', remove = FALSE, convert = TRUE)
# Potreiro Code number word
#1 7A 5 5
#2 7A 2 2
#3 7A ad NA ad
#4 7A 4 4
#5 7A 3 3
#6 7B 5 5
#7 7A 70% 70
#8 7B 50% 50
#9 7A ad NA ad
#10 7B ac NA ac
非常感谢。那很有用谢谢你!那是helpful@Júlio Azambuja是否需要将百分比放在单独的列中?@Júlio Azambuja是否需要将百分比放在单独的列中?
library(dplyr)
library(tidyr)
df %>%
extract(Code, into = c('number', 'word'), '(\\d*)([a-z]*)', remove = FALSE, convert = TRUE)
# Potreiro Code number word
#1 7A 5 5
#2 7A 2 2
#3 7A ad NA ad
#4 7A 4 4
#5 7A 3 3
#6 7B 5 5
#7 7A 70% 70
#8 7B 50% 50
#9 7A ad NA ad
#10 7B ac NA ac